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Algebra and Graphs - Linear, quadratic, and exponential graphs

Grade 11IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Linear Graphs: Represented by the equation y=mx+cy = mx + c, where mm is the gradient and cc is the y-intercept. These are straight lines.

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Quadratic Graphs: Represented by y=ax2+bx+cy = ax^2 + bx + c. They form a parabola (U-shaped if a>0a > 0, n-shaped if a<0a < 0). Key features include the vertex (turning point) and the axis of symmetry.

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Exponential Graphs: Represented by y=axy = a^x or y=k⋅axy = k \cdot a^x. These graphs show rapid growth (if a>1a > 1) or decay (if 0<a<10 < a < 1) and usually have the x-axis (y=0y=0) as a horizontal asymptote.

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Intersections: The points where two graphs cross represent the simultaneous solutions to their equations.

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Gradient Calculation: The steepness of a line, calculated as the vertical change divided by the horizontal change between two points.

📐Formulae

m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1} (Gradient formula)

y=mx+cy = mx + c (Slope-intercept form)

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} (Quadratic formula to find x-intercepts)

x=−b2ax = -\frac{b}{2a} (X-coordinate of the vertex of a quadratic)

m1⋅m2=−1m_1 \cdot m_2 = -1 (Condition for perpendicular lines)

💡Examples

Problem 1:

Find the equation of the line passing through the points (2,5)(2, 5) and (4,9)(4, 9).

Solution:

y=2x+1y = 2x + 1

Explanation:

First, calculate the gradient: m=(9−5)/(4−2)=4/2=2m = (9 - 5) / (4 - 2) = 4 / 2 = 2. Substitute m=2m=2 and point (2,5)(2, 5) into y=mx+cy = mx + c: 5=2(2)+c⇒5=4+c⇒c=15 = 2(2) + c \Rightarrow 5 = 4 + c \Rightarrow c = 1. Thus, y=2x+1y = 2x + 1.

Problem 2:

Find the coordinates of the turning point for the quadratic graph y=x2−6x+5y = x^2 - 6x + 5.

Solution:

(3,−4)(3, -4)

Explanation:

The x-coordinate of the vertex is x=−b/(2a)=−(−6)/(2⋅1)=3x = -b / (2a) = -(-6) / (2 \cdot 1) = 3. Substitute x=3x = 3 back into the original equation to find yy: y=(3)2−6(3)+5=9−18+5=−4y = (3)^2 - 6(3) + 5 = 9 - 18 + 5 = -4.

Problem 3:

Determine the value of kk for the exponential graph y=k⋅2xy = k \cdot 2^x if it passes through the point (3,40)(3, 40).

Solution:

k=5k = 5

Explanation:

Substitute the coordinates (x,y)=(3,40)(x, y) = (3, 40) into the equation: 40=k⋅2340 = k \cdot 2^3. Since 23=82^3 = 8, the equation becomes 40=8k40 = 8k. Dividing both sides by 8 gives k=5k = 5.